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There is 56 ounces of green paint in total. 75% is a batch of blue paint and 25% is yellow. To make a new batch of green paint,

Alecsey [184]

Answer:

60%

Step-by-step explanation:

There is 56 total ounces of green paint.

42oz is = 3/4

14oz is = 1/4

If she doubles the amount of yellow paint it would be 28oz instead of 14.

42+28= 70

There is now 70oz of paint total which becomes the new denominator.

The numerator is 42oz because that is how much blue paint there is.

42/70=.6 or 60%

6 0

3 years ago

A record store owner finds that 20% of customers entering her store make a purchase. One morning 180 people, who can be regarded

katen-ka-za [31]

Answer:

a) 0.2

b) 0.0009

c) 0.0298

d) 0.0465 = 4.65% probability that the sample proportion is less than 0.15.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$